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Current Question (ID: 7522)

Question:

$\text{Be}^{3+} \text{ ion and a proton are accelerated by the same potential. The ratio of their de-Broglie wavelengths is}$\n\n$\text{(assume mass of proton = mass of neutron)}$

Options:

1. $1:2$
2. $1:4$
3. $1:1$
4. $1:3\sqrt{3}$

Solution:

$\text{The de-Broglie wavelength formula is } \lambda = \frac{h}{mv} \text{, where } h \text{ is Planck's constant, } m \text{ is mass, and } v \text{ is velocity.}$\n\n$\text{For a particle accelerated through a potential difference } V \text{, the kinetic energy gained is } K = qV \text{, where } q \text{ is the charge.}$\n\n$\text{Using } K = \frac{1}{2}mv^2 \text{, we can express the velocity as } v = \sqrt{\frac{2qV}{m}}$\n\n$\text{Therefore, the de-Broglie wavelength can be written as:}$\n\n$\lambda = \frac{h}{m\sqrt{\frac{2qV}{m}}} = \frac{h}{\sqrt{2qVm}}$\n\n$\text{For a proton:}$\n\n$\lambda_p = \frac{h}{\sqrt{2eVm_p}}$\n\n$\text{For Be}^{3+} \text{ion:}$\n\n$\lambda_{\text{Be}^{3+}} = \frac{h}{\sqrt{2 \times 3eVm_{\text{Be}^{3+}}}}$\n\n$\text{The mass of Be}^{3+} \text{ ion is approximately } 9m_p \text{ (Be has mass number 9):}$\n\n$\lambda_{\text{Be}^{3+}} = \frac{h}{\sqrt{2 \times 3eV \times 9m_p}} = \frac{h}{\sqrt{54eVm_p}}$\n\n$\text{Now, let's find the ratio:}$\n\n$\frac{\lambda_{\text{Be}^{3+}}}{\lambda_p} = \frac{h/\sqrt{54eVm_p}}{h/\sqrt{2eVm_p}} = \sqrt{\frac{2eVm_p}{54eVm_p}} = \sqrt{\frac{2}{54}} = \sqrt{\frac{1}{27}} = \frac{1}{3\sqrt{3}}$\n\n$\text{Therefore, the ratio of wavelengths } \lambda_{\text{Be}^{3+}}:\lambda_p \text{ is } 1:3\sqrt{3}$

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}